Vector Calculus: Flux and divergence theorem

**bobmarley909** · May 19th 2010, 01:20 PM

1. Consider a cube with vertices at A=(0,0,0) B=(2,0,0) C=(2,2,0) D=(0,2,0) E=(0,0,2) F=(2,0,2) G=(2,2,2) H=(0,2,2)
A)Calculate the flux of the vector fieldF=xi through each face of the cube by taking the normal vectors pointing outwards.
B)Verify Gauss's divergence theorem for the cube and the vector field F by computing each side of the formula.
C)Using Gauss's divergence theorem evaluate the flux $\int \int F.dS$ of the vector field F=xi+yj+ $z^2$ k where S is a closed surface consisting of the cylinder $x^2 + y^2 = a^2$ , 0<z<b and the circular disks $x^2 + y^2 , a^2$ at z=0 and $x^2 + y^2 = a^2$ at z=b.

I have the basics but dont know how to get an answer. My workings are attatched.

**11rdc11** · May 19th 2010, 02:08 PM

Originally Posted by bobmarley909

1. Consider a cube with vertices at A=(0,0,0) B=(2,0,0) C=(2,2,0) D=(0,2,0) E=(0,0,2) F=(2,0,2) G=(2,2,2) H=(0,2,2)
A)Calculate the flux of the vector fieldF=xi through each face of the cube by taking the normal vectors pointing outwards.
B)Verify Gauss's divergence theorem for the cube and the vector field F by computing each side of the formula.
C)Using Gauss's divergence theorem evaluate the flux $\int \int F.dS$ of the vector field F=xi+yj+ $z^2$ k where S is a closed surface consisting of the cylinder $x^2 + y^2 = a^2$ , 0<z<b and the circular disks $x^2 + y^2 , a^2$ at z=0 and $x^2 + y^2 = a^2$ at z=b.

I have the basics but dont know how to get an answer. My workings are attatched.

$z= 0 ,n= -k ,F ^. n = 0$

$z = 2, n= k, F^.n =0$

$y = 0, n= -j, F^.n = 0$

$y = 2, n=j, F^. n = 0$

$x=0, n=-i, F^. n = 0$

$x = 2, n = i, F^. n =2$

So now

$\int^{2}_{0} \int^{2}_{0} 2 dxdy = 8$

Using divergence theorem

$\int^{2}_{0} \int^{2}_{0}\int^{2}_{0} dxdydz = 8$

c)

$\int^{2\pi}_{0} \int^{a}_{0} \int^{b}_{0} (2 +2z)r dzdrd\theta$

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